InterviewSolution
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InterviewSolution
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Evaluate : (i) `intsqrt(x^(2)-16)dx` (ii) `intsqrt(4x^(2)-5)dx` (iii) `intsqrt(17x^(2)-11)dx` |
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Answer» We know that `intsqrt(x^2-a^(2))dx=(x)/(2)sqrt(x^(2)-a^(2))-(a^(2))/(2)log|x+sqrt(x^(2)-a^(2))|+C`. `therefore` (i) `sqrt(x^(2)-16)dx=intsqrt(x^(2)-4^(2))dx` `=(x)/(2)*sqrt(x^(2)-4^(2))-(16)/(2)log|x+sqrt(x^(2)-16)|+C` `=(x)/(2)*sqrt(x^(2)-16)-8log|x+sqrt(x^(2)-16)|+C`. (ii) `intsqrt(4x^(2)-5)dx=2intsqrt(x^(2)-(5)/(4))dx=2intsqrt(x^(2)-((sqrt(5))/(2))^(2))dx` `=2*[(x)/(2)sqrt(x^(2)-(5)/(4))-(5)/(8)-log|x+sqrt(x^(2)-(5)/(4))|]+C` `=xsqrt(x^(2)-(5)/(4))-(5)/(4)log|xsqrt(x^(2)-(5)/(4))|+C`. (iii) `intsqrt(17x^(2)-11)dx=sqrt(17)*intsqrt(x^(2)-(11)/(17))dx` `=sqrt(17)*{(x)/(2)sqrt(x^(2)-(11)/(17))-(11)/(34)log|x+sqrt(x^(2)-(11)/(17))|}+C` `=(x)/(2)sqrt(17x^(2)-11-)(11sqrt(17))/(34)log|x+sqrt(x^(2)-(11)/(17))|+C`. |
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